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bayley86 · Jan 7, 2010, 02:55 PM

reduce the following algerbraic fractions to partial fractions

7x-4
(x-2)(2x-3)

5x-7
(3x+2)^2

(x-3)(x+4)
x^2-12x
can anyone help me with these

thanks

Andy

galactus · Jan 7, 2010, 05:52 PM

Originally Posted by bayley86

reduce the following algerbraic fractions to partial fractions

7x-4
(x-2)(2x-3)

I will show you this one, then you try the others.

Rewrite as $\frac{A}{x-2}+\frac{B}{2x-3}=7x-4$

The idea is to find A and B.

$A(2x-3)+B(x-2)=7x-4$ ... [1]

What makes 2x-3=0? x=3/2. Sub this into [1]

$\frac{-1}{2}B=\frac{13}{2}$

$B=-13$

What makes x-2=0, x=2 of course. Sub this into [1]

$A(2(2)-3)=7(2)-4$

$A=10$

There it is:

$\fbox{\frac{10}{x-2}-\frac{13}{2x-3}}$

bayley86 · Jan 9, 2010, 05:14 AM

hi this is the bit I don't understand or were you got it from
-1 B=13
2 2
B=-13
please could you explain what I have to do step by step

thanks

Andy

Unknown008 · Jan 9, 2010, 10:00 AM

Well, this was a good explanation... I'll do deeper then. We started with

$\frac{7x-4}{(x-2)(2x-3)} = \frac{A}{x-2}+\frac{B}{2x-3}$

Multiplied everything by the denominator:

$A(2x-3)+B(x-2)=7x-4$

You need to have A removed completely to find B. That means A has to be multiplied by 0. In turn, that means that 2x-3 = 0, or x = 3/2.

Well, OK, replace all the 'x' by 3/2.

$A(2(\frac32)-3)+B((\frac32)-2)=7(\frac32)-4$

$A(3-3)+B(-\frac12)=\frac{21}2-4$

$A(0)+B(-\frac12)=\frac{13}2$

$-\frac12 B=\frac{13}2$

$B=-13$

Then, you find how to make B disappear by making x-2 = 0.

galactus · Jan 9, 2010, 10:17 AM

$\frac{(x-3)(x+4)}{x^{2}-12x}$

$\frac{A}{x}+\frac{B}{x-12}=(x-3)(x+4)$

$\frac{5x-7}{(3x+2)^{2}}$

$\frac{A}{3x+2}+\frac{B}{(3x+2)^{2}}=5x-7$